Question: The equation of a circle $C$ is $x^2+y^2-18x+4y+49 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2-18x) + (y^2+4y) = -49$ $(x^2-18x+81) + (y^2+4y+4) = -49 + 81 + 4$ $(x-9)^{2} + (y+2)^{2} = 36 = 6^2$ Thus, $(h, k) = (9, -2)$ and $r = 6$.